Bất phương trình \(\sqrt {x - 1} > \sqrt {x - 2} + \sqrt {x - 3} \) có bao nhiêu nghiệm nguyên dương?

* Hướng dẫn giải

ĐKXĐ: \(\left\{ \begin{array}{l}x - 1 \ge 0\\x - 2 \ge 0\\x - 3 \ge 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\x \ge 2\\x \ge 3\end{array} \right.\)\( \Leftrightarrow x \ge 3\)

\(\begin{array}{l}\sqrt {x - 1}  > \sqrt {x - 2}  + \sqrt {x - 3} \\ \Leftrightarrow x - 1 > x - 2 + x - 3 + 2\sqrt {\left( {x - 2} \right)\left( {x - 3} \right)} \\ \Leftrightarrow 2\sqrt {\left( {x - 2} \right)\left( {x - 3} \right)}  < 4 - x\\ \Leftrightarrow \left\{ \begin{array}{l}4 - x > 0\\\left( {x - 2} \right)\left( {x - 3} \right) \ge 0\\4\left( {x - 2} \right)\left( {x - 3} \right) < {\left( {4 - x} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x < 4\\\left[ \begin{array}{l}x \le 2\\x \ge 3\end{array} \right.\\4{x^2} - 20x + 24 < 16 - 8x + {x^2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x < 4\\\left[ \begin{array}{l}x \le 2\\x \ge 3\end{array} \right.\\3{x^2} - 12x + 8 < 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \le 2\\3 \le x < 4\end{array} \right.\\\dfrac{{6 - 2\sqrt 3 }}{3} < x < \dfrac{{6 + 2\sqrt 3 }}{3}\end{array} \right.\\ \Leftrightarrow x \in \left( {\dfrac{{6 - 2\sqrt 3 }}{3};2} \right] \cup \left[ {3;\dfrac{{6 - 2\sqrt 3 }}{3}} \right)\end{array}\)

Mà \(x \in {\mathbb{Z}^ + }\) và \(x \ge 3\) nên \(x = 3\).

Chọn B.